Question

The time between failures of a laser in a machine, X, is exponentially distributed with a mean of 25,000 hours. In other words, 1 a= (failures/hour). 25,000 Exponential Distribution (pdf): f(x) = 1.0-\x, for x > 0. (a) What is the probability that the next failure occurs in 27,000 hours? (b) What is the expected time until the third failure? (c) What is the probability that the time until the third failure exceeds 25,000 hours?

Answer 1

a) The probability that the next failure occurs in 27000 hours is computed here as:

$P(X < 27000) = \int_{0}^{27000}(1/25000)e^{-x/25000} \ dx = 1 - e^{-27000/25000} = 0.6604$

therefore 0.6604 is the required probability here.

b) The expected time until the third failure here is computed here as:

= 3*Expected time for one failure

= 3*25,000 = 75,000 hours

Therefore 75,000 hours is the expected time here.

c) As the mean time to failure is 25,000 hours, therefore the mean number of failures in 25,000 hours is given as 1.

The probability that the time until third failure is more than 25,000 hours is computed here as:

= Probability that there is 0,1 or 2 in 25,000 hours period

$= e^{-1} + e^{-1} + \frac{1}{2}e^{-1} = 0.9197$

Therefore 0.9197 is the required probability here.